Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct ops {
  string s;
  long long w, t;
};
string todo[] = {"row", "col"};
long long a[111][111], b[111][111];
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  if (n > m) {
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < i; j++) {
        swap(a[i][j], a[j][i]);
      }
    }
    swap(n, m);
    swap(todo[0], todo[1]);
  }
  long long sz = 0;
  vector<ops> ans;
  for (long long i = 0; i < n; i++) {
    long long mn = INT_MAX;
    for (long long j = 0; j < m; j++) {
      mn = min(a[i][j], mn);
    }
    sz += mn;
    if (mn) ans.push_back({todo[0], i + 1, mn});
    for (long long j = 0; j < m; j++) {
      b[i][j] = mn;
    }
  }
  for (long long j = 0; j < m; j++) {
    long long xtra = a[0][j] - b[0][j];
    sz += xtra;
    if (xtra) {
      ans.push_back({todo[1], j + 1, xtra});
    }
    for (long long i = 0; i < n; i++) {
      b[i][j] += xtra;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      if (a[i][j] != b[i][j]) {
        cout << "-1";
        return;
      }
    }
  }
  cout << sz << '\n';
  for (long long i = 0; i < ans.size(); i++) {
    for (long long j = 0; j < ans[i].t; j++) {
      cout << ans[i].s << ' ' << ans[i].w << '\n';
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}